Problem: Consider the polar curve $r=e^\theta$ for $0\le \theta < 2\pi$. At which values of $\theta$ does the graph of $r$ have a horizontal tangent line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3\pi}{4}$ only (Choice B) B $\dfrac{7\pi}{4}$ only (Choice C) C $\dfrac{3\pi}{4}$ or $\dfrac{7\pi}{4}$ (Choice D) D The graph of $r$ has no horizontal tangents.
Explanation: A horizontal line has a slope of $0$. The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Once we have an expression for the slope of the tangent line, we can look for the $\theta$ -values that make $\dfrac{dy}{d\theta}=0$ but don't make $\dfrac{dx}{d\theta}=0$ (because then $\dfrac{dy}{dx}$ will be undefined). For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={e^\theta}\cos(\theta) \\\\ y&={e^\theta} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$. $\begin{aligned} y(\theta)&=e^\theta\sin(\theta) \\\\ \dfrac{dy}{d\theta}&=e^\theta\sin(\theta)+e^\theta\cos(\theta) \\\\ &=e^\theta(\sin(\theta)+\cos(\theta)) \\\\ \\\\ x(\theta)&=e^\theta\cos(\theta) \\\\ \dfrac{dx}{d\theta}&=e^\theta\cos(\theta)+e^\theta(-\sin(\theta)) \\\\ &=e^\theta(\cos(\theta)-\sin(\theta)) \end{aligned}$ Now let's solve $\dfrac{dy}{d\theta}=0$ on the interval $0\le \theta < 2\pi$. $\begin{aligned} \dfrac{dy}{d\theta}&=0 \\\\ e^\theta(\sin(\theta)+\cos(\theta)) &=0 \\\\ \sin(\theta)+\cos(\theta)&=0 \\\\ \sin(\theta)&=-\cos(\theta) \end{aligned}$ Within our interval, our possible solutions are $\theta=\dfrac{3\pi}{4}$ and $\theta=\dfrac{7\pi}{4}$. Finally, we evaluate $\dfrac{dx}{d\theta}$ for our two possible values of $\theta$ and require that $\dfrac{dx}{d\theta}\ne0$. $\begin{aligned} \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{3\pi }{4}}}&=e^{^{\left({\tfrac{3\pi}{4}}\right)}}\left(\cos\left({\dfrac{3\pi}{4}}\right)-\sin\left({\dfrac{3\pi}{4}}\right)\right) \\\\ &=e^{^{\left({\tfrac{3\pi}{4}}\right)}}\left(-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\right) \\\\ &=-\sqrt{2} e^{^{\left({\tfrac{3\pi}{4}}\right)}} \\\\ \\\\ \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{7\pi }{4}}}&=e^{^{\left({\tfrac{7\pi}{4}}\right)}}\left(\cos\left({\dfrac{7\pi}{4}}\right)-\sin\left({\dfrac{7\pi}{4}}\right)\right) \\\\ &=e^{^{\left({\tfrac{7\pi}{4}}\right)}}\left(\dfrac{\sqrt{2}}{2}-\left(-\dfrac{\sqrt{2}}{2}\right)\right) \\\\ &=\sqrt{2}e^{^{\left({\tfrac{7\pi}{4}}\right)}} \end{aligned}$ The graph of $r$ has a horizontal tangent at $\theta=\dfrac{3\pi}{4}$ or $\theta=\dfrac{7\pi}{4}$. The graphs of the tangents are shown with the figure at different scales. ${3}$ ${6}$ ${9}$ ${12}$ ${0}$ ${\frac{1}{8}\pi}$ ${\frac{1}{4}\pi}$ ${\frac{3}{8}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{5}{8}\pi}$ ${\frac{3}{4}\pi}$ ${\frac{7}{8}\pi}$ ${\pi}$ ${\frac{9}{8}\pi}$ ${\frac{5}{4}\pi}$ ${\frac{11}{8}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{13}{8}\pi}$ ${\frac{7}{4}\pi}$ ${\frac{15}{8}\pi}$ ${100}$ ${200}$ ${0}$ ${\frac{1}{8}\pi}$ ${\frac{1}{4}\pi}$ ${\frac{3}{8}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{5}{8}\pi}$ ${\frac{3}{4}\pi}$ ${\frac{7}{8}\pi}$ ${\pi}$ ${\frac{9}{8}\pi}$ ${\frac{5}{4}\pi}$ ${\frac{11}{8}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{13}{8}\pi}$ ${\frac{7}{4}\pi}$ ${\frac{15}{8}\pi}$